Question

clock in the moving coordinate system reads t’ = 0 when the stationary clock reads t = 0. If the moving frame moves at a speed of 0.800c, what time will the moving clock read when the stationary observer reads 15.0 hr on her clock?

Answer 1

We know that if one frame is moving with a speed v = 0.800c, then the time between the two will vary by a factor of

$\sqrt{(1-v^{2}/c^{2})}= \sqrt{(1-(0.800c)^{2}/c^{2})}=\sqrt{(1-0.800^{2})}=0.6$

Thus, the time measured in system accelerating will be 0.6 times of the time measured in system at rest.

So, the time measured in accelerating system is

$t^{'}=0.6\times 15=9hr$